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The tree is represented by a list, with the use of heapq moddule. • First, create a node use each (key, value) pair of the frequency dict. • Then, use heapq.heappush() to insert all the nodes to the tree list, while maintaining the minimum heap invariant, that is, heap[k] <= heap[2*k+1] and heap[k] <= heap[2*k+2] . See the document for more information: https://docs.python.org/3/library/heapq.html I 1: def create_tree(frequency): tree = [] for key, val in frequency.items(): ### START YOUR CODE ### node = None # Create a node pass # Insert the node to tree ### END YOUR CODE ### return tree I ]: # Do not change the test code here sample_text = 'No, it is a word. What matters is the connection the word implies.' freq = create_frequency_dict(sample_text) %3D tree = create_tree(freq)

The tree is represented by a list, with the use of heapq moddule. • First, create a node use each (key, value) pair of the frequency dict. • Then, use heapq.heappush() to insert all the nodes to the tree list, while maintaining the minimum heap invariant, that is, heap[k] <= heap[2*k+1] and heap[k] <= heap[2*k+2] . See the document for more information: https://docs.python.org/3/library/heapq.html I 1: def create_tree(frequency): tree = [] for key, val in frequency.items(): ### START YOUR CODE ### node = None # Create a node pass # Insert the node to tree ### END YOUR CODE ### return tree I ]: # Do not change the test code here sample_text = 'No, it is a word. What matters is the connection the word implies.' freq = create_frequency_dict(sample_text) %3D tree = create_tree(freq)

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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The tree is represented by a list, with the use of heapq moddule. • First, create a node use each (key, value) pair of the frequency dict. Then, use heapq.heappush() to insert all the nodes to the tree list, while maintaining the minimum heap invariant, that is, heap[k] <= heap[2*k+1] and heap[k] <= heap[2*k+2] . See the document for more information: https://docs.python.org/3/library/heapg.html In [ ]: def create_tree(frequency): tree = [] for key, val in frequency.items (): ### START YOUR CODE ### node = None # Create a node pass # Insert the node to tree ### END YOUR CODE ### return tree In [ 1: # Do not change the test code here sample text = 'No, it is a word. What matters is the connection the word implies.' freq = create_frequency_dict(sample_text) tree = create_tree(freq) print(sorted(tree, key=lambda x: x.freq) [:5]) Expected output: [Node(N, 1), Node(p, 1), Node(., 1), Node(l, 1), Node(W, 1]
Transcribed Image Text:The tree is represented by a list, with the use of heapq moddule. • First, create a node use each (key, value) pair of the frequency dict. Then, use heapq.heappush() to insert all the nodes to the tree list, while maintaining the minimum heap invariant, that is, heap[k] <= heap[2*k+1] and heap[k] <= heap[2*k+2] . See the document for more information: https://docs.python.org/3/library/heapg.html In [ ]: def create_tree(frequency): tree = [] for key, val in frequency.items (): ### START YOUR CODE ### node = None # Create a node pass # Insert the node to tree ### END YOUR CODE ### return tree In [ 1: # Do not change the test code here sample text = 'No, it is a word. What matters is the connection the word implies.' freq = create_frequency_dict(sample_text) tree = create_tree(freq) print(sorted(tree, key=lambda x: x.freq) [:5]) Expected output: [Node(N, 1), Node(p, 1), Node(., 1), Node(l, 1), Node(W, 1]
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